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प्रश्न
उत्तर
\[\int\left( \frac{x + 1}{\sqrt{2x + 3}} \right)dx\]
\[ = \frac{1}{2}\int\left( \frac{2x + 2}{\sqrt{2x + 3}} \right)dx\]
\[ = \frac{1}{2}\int\left( \frac{2x + 3 - 1}{\sqrt{2x + 3}} \right)dx\]
\[ = \frac{1}{2}\int\left( \frac{2x + 3}{\sqrt{2x + 3}} - \frac{1}{\sqrt{2x + 3}} \right)dx\]
\[ = \frac{1}{2}\int\left( \sqrt{2x + 3} - \frac{1}{\sqrt{2x + 3}} \right)dx\]
\[ = \frac{1}{2}\left[ \int \left( 2x + 3 \right)^\frac{1}{2} dx - \int \left( 2x + 3 \right)^{- \frac{1}{2}} dx \right]\]
\[ = \frac{1}{2}\left[ \frac{\left( 2x + 3 \right)^\frac{1}{2} + 1}{2\left( \frac{1}{2} + 1 \right)} - \frac{\left( 2x + 3 \right)^{- \frac{1}{2} + 1}}{2\left( - \frac{1}{2} + 1 \right)} + C \right]\]
\[ = \frac{1}{2}\left[ \frac{1}{3} \left( 2x + 3 \right)^\frac{3}{2} - \left( 2x + 3 \right)^\frac{1}{2} + C \right]\]
\[ = \frac{1}{6} \left( 2x + 3 \right)^\frac{3}{2} - \frac{1}{2} \left( 2x + 3 \right)^\frac{1}{2} + C\]
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संबंधित प्रश्न
\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then
\[\int\frac{5 x^4 + 12 x^3 + 7 x^2}{x^2 + x} dx\]
